This is a clarification of another post of mine.

Fix $n$ a positive integer. Let $SL(n)$ have its usual matrix representation, so that it really is the codimension-one subset of $M(n) = \mathbb R^{n^2}$ cut out by the degree-$n$ condition that the determinant is $1$. So we have $n^2$ coordinate functions $A^i_j$ on $SL(n)$, $i,j = 1,\dots,n$.

Let $U$ be a domain in $\mathbb R^n$, with coordinates $x_1,\dots,x_n$. Consider the set $\mathcal S$ of smooth functions $f: U \to SL(n)$ satisfying the differential equation $\frac{\partial f^i_j}{\partial x^k} = \frac{\partial f^i_k}{\partial x^j}$ for each $i,j,k = 1,\dots,n$ (of course, $f^i_j = A^i_j \circ f$ is the $(i,j)$th coordinate of $f$).

(Why would you care about $\mathcal S$? Because a smooth map $g: U \to \mathbb R^n$ is volume-preserving if and only if $\frac{\partial g^i}{\partial x^j} \in \mathcal S$, and every element of $\mathcal S$ arises this way; indeed, $\mathcal S$ is the space of volume-preserving maps up to translations.)

Let's agree that a *smooth path* in $\mathcal S$ is a smooth function $F: [0,1] \times U \to SL(n)$ such that for each $t\in [0,1]$, $F(t,-) \in \mathcal S$.

**Question:** Is $\mathcal S$ smooth-path-connected? I.e. given $f_0, f_1 \in \mathcal S$, does there exist a smooth path $F$ so that $F(0,-) = f_0$ and $F(1,-) = f_1$?

If the answer is "no" in general, is it "yes" for sufficiently nice domains $U$ (contractible, say, or with compact closure and require that each $f\in \mathcal S$ extend smoothly to a neighborhood of the closure, or...)?